We know that the roots of x^7 + 1 = 0 are given by :
x(t) = e^[i*(2t-1)*pi/7] for t = 1,2,3,4,5,6,7

Since the coefficient of x^6 in the given equation
is 0, it follows that:

Sum( i = 1 To 7)[e^[i*(2t-1)*pi/7]] = 0

This is possible iff the respective real parts of e^[(2t-1)*pi/2]
which are cos[(2t-1)*pi/7] for t = 1, 2, ..., 7 sum to 0.

Remembering that cos(7*pi/7) = -1 in conjunction with the identity:
cos(2*pi - x) = cos x, this gives:

2[cos (pi/7) + cos (3pi/7) + cos (pi/7)] = 1
Or, cos (pi/7) + cos (3pi/7) + cos (pi/7) = 1/2

Now, cos(5*pi/7) = cos(pi - 2*pi/7) = - cos(2*pi/7), and consequently:
cos pi/7 - cos(2*pi/7) + cos (3*pi/7) = 1/2